I am new to ruby and I want to do the following action to remove last "_val3" in ruby:
$ val="val1_val2_val3"
$ echo ${val%_*}
val1_val2
I used to use echo ${val%_*} to get "val1_val2", but i do not how do this in ruby.
Also, how to get "val1"?
Is there a good way to do them?
Not a ruby expert but I'll get the ball rolling with a regular expression:
a.sub /_[^_]*$/, ''
Match an underscore followed by any number of non-underscores at the end of the string. Replace with nothing.
You can use a single gsub to get your expected o/p,
a = "a-b_c_d"
# => "a-b_c_d"
a.gsub /_[a-z]*$/, ''
# => "a-b_c"
Or, you can use ruby split and join,
a.split("_")[0..-2].join("_")
# => "a-b_c"
String#rpartition would probably work:
'a-b_c_d'.rpartition('_') #=> ["a-b_c", "_", "d"]
rpartition looks for the last '_' and returns an array containing the part before it, the separator itself and the part after it.
Related
How can I get the username without the # symbol?
That's everything between # and any non-word character.
message = <<-MESSAGE
From #victor with love,
To #andrea,
and CC goes to #ghost
MESSAGE
Using a Ruby regular expression, I tried
username_pattern = /#\w+/
I will like to get the following output
message.scan(username_pattern)
#=> ["victor", "andrea", "ghost"]
Use look behind
(?<=#)\w+
this will leave # symbol regex
I would go with:
message.scan(/(?<=#)\w+/)
#=> ["victor","andrea","ghost"]
You might want to read about look-behind regexp.
You could match the # and then capture one or more times a word character in a capturing group
#(\w+)
username_pattern = /#(\w+)/
Regex demo
Try this
irb(main):010:0> message.scan(/#(\w+)/m).flatten
=> ["victor", "andrea", "ghost"]
I am trying to use gsub or sub on a regex passed through terminal to ARGV[].
Query in terminal: $ruby script.rb input.json "\[\{\"src\"\:\"
Input file first 2 lines:
[{
"src":"http://something.com",
"label":"FOO.jpg","name":"FOO",
"srcName":"FOO.jpg"
}]
[{
"src":"http://something123.com",
"label":"FOO123.jpg",
"name":"FOO123",
"srcName":"FOO123.jpg"
}]
script.rb:
dir = File.dirname(ARGV[0])
output = File.new(dir + "/output_" + Time.now.strftime("%H_%M_%S") + ".json", "w")
open(ARGV[0]).each do |x|
x = x.sub(ARGV[1]),'')
output.puts(x) if !x.nil?
end
output.close
This is very basic stuff really, but I am not quite sure on how to do this. I tried:
Regexp.escape with this pattern: [{"src":".
Escaping the characters and not escaping.
Wrapping the pattern between quotes and not wrapping.
Meditate on this:
I wrote a little script containing:
puts ARGV[0].class
puts ARGV[1].class
and saved it to disk, then ran it using:
ruby ~/Desktop/tests/test.rb foo /abc/
which returned:
String
String
The documentation says:
The pattern is typically a Regexp; if given as a String, any regular expression metacharacters it contains will be interpreted literally, e.g. '\d' will match a backlash followed by ‘d’, instead of a digit.
That means that the regular expression, though it appears to be a regex, it isn't, it's a string because ARGV only can return strings because the command-line can only contain strings.
When we pass a string into sub, Ruby recognizes it's not a regular expression, so it treats it as a literal string. Here's the difference in action:
'foo'.sub('/o/', '') # => "foo"
'foo'.sub(/o/, '') # => "fo"
The first can't find "/o/" in "foo" so nothing changes. It can find /o/ though and returns the result after replacing the two "o".
Another way of looking at it is:
'foo'.match('/o/') # => nil
'foo'.match(/o/) # => #<MatchData "o">
where match finds nothing for the string but can find a hit for /o/.
And all that leads to what's happening in your code. Because sub is being passed a string, it's trying to do a literal match for the regex, and won't be able to find it. You need to change the code to:
sub(Regexp.new(ARGV[1]), '')
but that's not all that has to change. Regexp.new(...) will convert what's passed in into a regular expression, but if you're passing in '/o/' the resulting regular expression will be:
Regexp.new('/o/') # => /\/o\//
which is probably not what you want:
'foo'.match(/\/o\//) # => nil
Instead you want:
Regexp.new('o') # => /o/
'foo'.match(/o/) # => #<MatchData "o">
So, besides changing your code, you'll need to make sure that what you pass in is a valid expression, minus any leading and trailing /.
Based on this answer in the thread Convert a string to regular expression ruby, you should use
x = x.sub(/#{ARGV[1]}/,'')
I tested it with this file (test.rb):
puts "You should not see any number [0123456789].".gsub(/#{ARGV[0]}/,'')
I called the file like so:
ruby test.rb "\d+"
# => You should not see any number [].
In Python language I find rstr that can generate a string for a regex pattern.
Or in Python we have this method that can return range of string:
re.sre_parse.parse(pattern)
#..... ('range', (97, 122)) ....
But In Ruby I didn't find any thing.
So how to generate string for a regex pattern in Ruby(reverse regex)?
I wanna to some thing like this:
"/[a-z0-9]+/".example
#tvvd
"/[a-z0-9]+/".example
#yt
"/[a-z0-9]+/".example
#bgdf6
"/[a-z0-9]+/".example
#564fb
"/[a-z0-9]+/" is my input.
The outputs must be correct string that available in my regex pattern.
Here outputs were: tvvd , yt , bgdf6 , 564fb that "example" method generated them.
I need that method.
Thanks for your advice.
You can also use the Faker gem https://github.com/stympy/faker and then use this call:
Faker::Base.regexify(/[a-z0-9]{10}/)
In Ruby:
/qweqwe/.to_s
# => "(?-mix:qweqwe)"
When you declare a Regexp, you've got the Regexp class object, to convert it to String class object, you may use Regexp's method #to_s. During conversion the special fields will be expanded, as you may see in the example., using:
(using the (?opts:source) notation. This string can be fed back in to Regexp::new to a regular expression with the same semantics as the original.
Also, you can use Regexp's method #inspect, which:
produces a generally more readable version of rxp.
/ab+c/ix.inspect #=> "/ab+c/ix"
Note: that the above methods are only use for plain conversion Regexp into String, and in order to match or select set of string onto an other one, we use other methods. For example, if you have a sourse array (or string, which you wish to split with #split method), you can grep it, and get result array:
array = "test,ab,yr,OO".split( ',' )
# => ['test', 'ab', 'yr', 'OO']
array = array.grep /[a-z]/
# => ["test", "ab", "yr"]
And then convert the array into string as:
array.join(',')
# => "test,ab,yr"
Or just use #scan method, with slightly changed regexp:
"test,ab,yr,OO".scan( /[a-z]+/ )
# => ["test", "ab", "yr"]
However, if you really need a random string matched the regexp, you have to write your own method, please refer to the post, or use ruby-string-random library. The library:
generates a random string based on Regexp syntax or Patterns.
And the code will be like to the following:
pattern = '[aw-zX][123]'
result = StringRandom.random_regex(pattern)
A bit late to the party, but - originally inspired by this stackoverflow thread - I have created a powerful ruby gem which solves the original problem:
https://github.com/tom-lord/regexp-examples
/this|is|awesome/.examples #=> ['this', 'is', 'awesome']
/https?:\/\/(www\.)?github\.com/.examples #=> ['http://github.com', 'http://www.github.com', 'https://github.com', 'https://www.github.com']
UPDATE: Now regular expressions supported in string_pattern gem and it is 30 times faster than other gems
require 'string_pattern'
/[a-z0-9]+/.generate
To see a comparison of speed https://repl.it/#tcblues/Comparison-generating-random-string-from-regular-expression
I created a simple way to generate strings using a pattern without the mess of regular expressions, take a look at the string_pattern gem project: https://github.com/MarioRuiz/string_pattern
To install it: gem install string_pattern
This is an example of use:
# four characters. optional: capitals and numbers, required: lower
"4:XN/x/".gen # aaaa, FF9b, j4em, asdf, ADFt
Maybe you can find what you are looking for over here.
In other languages, in RegExp you can use /.../g for a global match.
However, in Ruby:
"hello hello".match /(hello)/
Only captures one hello.
How do I capture all hellos?
You can use the scan method. The scan method will either give you an array of all the matches or, if you pass it a block, pass each match to the block.
"hello1 hello2".scan(/(hello\d+)/) # => [["hello1"], ["hello2"]]
"hello1 hello2".scan(/(hello\d+)/).each do|m|
puts m
end
I've written about this method, you can read about it here near the end of the article.
Here's a tip for anyone looking for a way to replace all regex matches with something else.
Rather than the //g flag and one substitution method like many other languages, Ruby uses two different methods instead.
# .sub — Replace the first
"ABABA".sub(/B/, '') # AABA
# .gsub — Replace all
"ABABA".gsub(/B/, '') # AAA
use String#scan. It will return an array of each match, or you can pass a block and it will be called with each match.
All the details at http://ruby-doc.org/core/classes/String.html#M000812
I have the following line
'passenger (2.2.5, 2.0.6)'.match(//)[0]
which obviously doesn't match anything yet
I want to return the just the content of (2.2.5, so everything after the open parentheses and before the comma.
How would I do this?
Beanish solution fails on more than 2 version numbers, you should use something like:
>> 'passenger (2.2.5, 2.0.6, 1.8.6)'.match(/\((.*?),/)[1] # => "2.2.5"
'passenger (2.2.5, 2.0.6)'.match(/\((.*),/)[1]
if you use the $1 element it is the group that is found within the ( )
#!/usr/bin/env ruby
s = 'passenger (2.2.5, 2.0.6)'
p s.scan(/(?:\(|, *)([^,)]*)/).flatten # => ["2.2.5", "2.0.6"]